Let $f(x)=\dfrac{1}{x^2-4}$ Where does $f$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-2$ (Choice B) B $x=0$ (Choice C) C $x=2$ (Choice D) D $f$ has no critical points.
Solution: A critical point of $f$ is a point in the domain of $f$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $f$, let's find its derivative. $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[ \dfrac{1}{x^2-4} \right] \\\\ &=-\left((x^2-4)^{-2}\right) \cdot \dfrac{d}{dx}\left[x^2-4\right] \\\\ &=-\dfrac{2x}{(x^2-4)^2} \end{aligned}$ Now let's look for $x$ -values where $f'$ is zero or undefined. $-\dfrac{2x}{(x^2-4)^2}=0$ at $x=0$. $-\dfrac{2x}{(x^2-4)^2}$ is undefined at $x=2$ and $x=-2$. However, $f(x)=\dfrac{1}{x^2-4}$ is also undefined at $x=2$ and $x=-2$ so they aren't critical points. In conclusion, this is the only $x$ -value where $f$ has a critical point: $x=0$